Math Is Fun Forum

denis_gylaev

Member

Registered: 2015-03-19

Posts: 66

Square problem

https://gyazo.com/c3c0e9e2aa6287ad65654b7965e503d4

so since each side as length one, and M and N are the midpoints, I used pythagoras to find the length of MD and BN which is 1.11803 but I don't know how that helps me

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Square problem

Make use of symmetry. The shaded area is congruent to unshaded area of parallelogram MBND. You can easily calculate area of MBND. Oral calculation tells you that required area is 1/4.

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{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

PCHydrogen

Member

Registered: 2016-03-09

Posts: 2

Re: Square problem

As it is symmetrical, you have your length MD 1.11803399
Turn it into a parallelogram, and as the area of a paralellogram is base * height, you already have the base as 1/2 is 0.5
0.5 * MD, which is the height, gets you 0.559016994.
0.559016994/2, as there are 2 identical trapeziums gets you 0.279508497.
Thus, the area of the shaded region is 0.279508497.

thickhead

Member

Registered: 2016-04-16

Posts: 1,086

Re: Square problem

Hi PCHydrogen,
The height of parallelogram is 1, Not the slant height.

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{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}